In integration by observation, we look at a product or a quotient to see a certain mathematical connection between its two parts.

**Case 1**

∫ f ^{'}(x) / f(x) dx = ln|f(x)| + c

If the **numerator** of the fraction is the differential coefficient of the **denominator**, then the **integral** is **ln|denominator|**

**Proof:**

∫ f '(x) / f(x) dx

let u = f(x) => du/dx = f '(x) => du = f '(x)dx

∫ f '(x) / f(x) dx = ∫ 1/u du

= ln|u| + c

= ln|f(x)| +c

**E.g.1**

∫ cot x dx

= ∫ cos x/sin x

= ln|sin x| + c

**E.g.2**

∫ tan x dx

= ∫ sin x/cos x

= - ∫ -sin x/cos x

= - ln|cos x| + c

or 0 - ln|cos x| = ln 1 - ln|cos x| = ln|1/cos x| = ln|sec x| + c

**E.g.3**

∫ x+1/x^{2} + 2x + 3 dx

= 1/2 ∫ 2x+2/x^{2} + 2x + 3 dx

= 1/2 ln|x^{2} + 2x + 3| + c

**E.g.4**

∫ 1/tan x cos^{2}x dx

= ∫ sec^{2} x/tan x dx

= ln|tan x| + c

**Case 2**

∫ f '(x) / [f(x)]^{n} dx = [f(x)]^{n+1}/n+1 + c

Please note that f '(x) is the differential coefficient of just the **inner function**, not the whole function.

**Proof:**

∫ f '(x) [f(x)]^{n} dx

let u = f(x) => du/dx = f '(x) => du = f '(x) dx

∫ f '(x) / [f(x)]^{n} dx = ∫ u^{n} du

= u^{n+1}/n+1 + c

= [f(x)]^{n+1}/n+1 + c

**E.g.1**

∫ sec^{2} x tan^{6} x dx

f '(x) = sec^{2} x and f(x) = tan x

∫ sec^{2} x tan^{6} x dx

= tan^{7}x / 7 + c

**E.g.2**

∫ (x + 1) (x^{2} + 2x -3) dx

= 1/2 ∫ (2x + 2) (x^{2} + 2x -3) dx

= (x^{2} + 2x -3)^{2}/2 + c

**E.g.3**

∫ (x - 1) √ (x^{2} - 2x) dx

= 1/2 ∫ (2x - 2) √ (x^{2} - 2x) dx

= 1/2 (x^{2} - 2x)^{3/2}/3/2 + c
= (x^{2} - 2x)^{3/2}/3 + c

**E.g.3**

∫ cos^{3}x dx

= ∫ cos x cos^{2} x dx

= ∫ cos x (1 - sin^{2}x) dx

= ∫ (cos x - cos x sin^{2}x) dx

= sin x - sin^{3}x / 3 + c

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